Count on every other line


 
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# 1  
Old 07-17-2008
Count on every other line

I'd like to create a loop that will display something like:
1
2
29
2
57
2
...
25173
2
I figure I'd want to make some code that counts to 1798 and for the odd numbers displays: 1+28((n-1)/2) and for the even numbers displays 2. This is what I wrote:
Code:
#!  /bin/csh
#include <stdio.h>

int count;

main()
{
  for( count = 1; count <1798; count++ )
    if (count % 2 == 1)
      printf ("\n%d", 1 + 28 * ( (count - 1) /2);
    if (count % 2 == 0)
      printf (2);;
    return 0;  
}

it didn't work though. Any idea why?
# 2  
Old 07-18-2008
Will re-post later

Last edited by Vi-Curious; 07-19-2008 at 12:49 AM.. Reason: Didn't pay attention to shell
# 3  
Old 07-19-2008
C shell is not something I ever use but, just quickly throwing something together, the following was done to keep it in a similar form with yours but it could easily be improved. Also, to get the output you showed, you would have to increase the upper limit of your loop.

Code:
 
#!/usr/bin/csh
set count = 1
while ($count <= 1800)
  if ($count % 2 == 1) then
    @ tmpval = (1 + 28 * (($count - 1) / 2))
    printf "%d\n" $tmpval
  endif
  if ($count % 2 == 0) \
    printf "2\n"
  @ count++
end

# 4  
Old 07-19-2008
Vi-Curious,

It's not a csh script but a c program.

Try this:

Code:
#include <stdio.h>

main()
{
  int count;
  
  for( count = 1; count < 1798; count++ ){
    if (count % 2 == 1) {
      printf ("%d\n", 1 + 28 * ((count - 1) /2));
    }
    if (count % 2 == 0) {
      printf ("%d\n", 2);
    }
  }
  return 0;  
}

Regards
# 5  
Old 07-19-2008
My first post assumed it was a program and it was obvious that, aside from a missing paren, the only real problem was the second printf. It was only after that that I noticed his first line identified it as a c shell. What I wouldn't give to be able to read minds. Smilie
 
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